Difference between revisions of "2008 AMC 10B Problems/Problem 10"

(New page: ==Problem== {{problem}} ==Solution== {{solution}} ==See also== {{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}})
 
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==Problem==
 
==Problem==
{{problem}}
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Suppose that <math>(u_n)</math> is a serquence of real numbers satifying <math>u_{n+2}=2u_{n+1}+u_n</math>,
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and that <math>u_3=9</math> and <math>u_6=128</math>. What is <math>u_5</math>?
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(A) 40 (B) 53 (C) 68 (D) 88 (E) 104
  
 
==Solution==
 
==Solution==
{{solution}}
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We know that <math>u_6=128</math>, so we plug in <math>n=4</math> to get <math>128=2u_5+u_4</math>. We plug in <math>n=3</math> to get
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<math>u_5=2u_4+9</math>. Substituting gives
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<math>128=5u_4+18 \rightarrow u_4=22</math>
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This gives <math>u_5=\frac{128-22}{2}=53</math>.
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Answer B is the correct answer
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NOTE: This is my (BOGTRO) solution, not the official one,
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and should be ignored in view of a better solution.
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}}
 
{{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}}

Revision as of 17:26, 10 August 2008

Problem

Suppose that $(u_n)$ is a serquence of real numbers satifying $u_{n+2}=2u_{n+1}+u_n$,

and that $u_3=9$ and $u_6=128$. What is $u_5$?

(A) 40 (B) 53 (C) 68 (D) 88 (E) 104

Solution

We know that $u_6=128$, so we plug in $n=4$ to get $128=2u_5+u_4$. We plug in $n=3$ to get

$u_5=2u_4+9$. Substituting gives

$128=5u_4+18 \rightarrow u_4=22$

This gives $u_5=\frac{128-22}{2}=53$.

Answer B is the correct answer

NOTE: This is my (BOGTRO) solution, not the official one,

and should be ignored in view of a better solution.

See also

2008 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions
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